Is a Rational Rotation Algebra a Cutdown of a Matrix Algebra?












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Let $theta=m/n$ and let $A_{theta}$ be the rational rotation C$^{*}$-algebra with rotation angle $theta$. I.e., $A_{theta}=C^{*}(u,v)$, where $u$ and $v$ are unitaries such that $vu=e^{2pi i theta}uv$. I know from here that $A_{theta}$ is not a full matrix algebra. Given that $A_{theta}$ has irreducible representations only of degree $n$, is it true that $A_{theta}$ is a cutdown of a full matrix algebra? I.e.,




Is there a space $C(X,M_{n}(mathbb{C}))$ and a projection $pin C(X,M_{n}(mathbb{C}))$, such that $A_{theta}$ is isomorphic to $pC(X,M_{n}(mathbb{C}))p$?











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    $begingroup$


    Let $theta=m/n$ and let $A_{theta}$ be the rational rotation C$^{*}$-algebra with rotation angle $theta$. I.e., $A_{theta}=C^{*}(u,v)$, where $u$ and $v$ are unitaries such that $vu=e^{2pi i theta}uv$. I know from here that $A_{theta}$ is not a full matrix algebra. Given that $A_{theta}$ has irreducible representations only of degree $n$, is it true that $A_{theta}$ is a cutdown of a full matrix algebra? I.e.,




    Is there a space $C(X,M_{n}(mathbb{C}))$ and a projection $pin C(X,M_{n}(mathbb{C}))$, such that $A_{theta}$ is isomorphic to $pC(X,M_{n}(mathbb{C}))p$?











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      $begingroup$


      Let $theta=m/n$ and let $A_{theta}$ be the rational rotation C$^{*}$-algebra with rotation angle $theta$. I.e., $A_{theta}=C^{*}(u,v)$, where $u$ and $v$ are unitaries such that $vu=e^{2pi i theta}uv$. I know from here that $A_{theta}$ is not a full matrix algebra. Given that $A_{theta}$ has irreducible representations only of degree $n$, is it true that $A_{theta}$ is a cutdown of a full matrix algebra? I.e.,




      Is there a space $C(X,M_{n}(mathbb{C}))$ and a projection $pin C(X,M_{n}(mathbb{C}))$, such that $A_{theta}$ is isomorphic to $pC(X,M_{n}(mathbb{C}))p$?











      share|cite|improve this question









      $endgroup$




      Let $theta=m/n$ and let $A_{theta}$ be the rational rotation C$^{*}$-algebra with rotation angle $theta$. I.e., $A_{theta}=C^{*}(u,v)$, where $u$ and $v$ are unitaries such that $vu=e^{2pi i theta}uv$. I know from here that $A_{theta}$ is not a full matrix algebra. Given that $A_{theta}$ has irreducible representations only of degree $n$, is it true that $A_{theta}$ is a cutdown of a full matrix algebra? I.e.,




      Is there a space $C(X,M_{n}(mathbb{C}))$ and a projection $pin C(X,M_{n}(mathbb{C}))$, such that $A_{theta}$ is isomorphic to $pC(X,M_{n}(mathbb{C}))p$?








      functional-analysis operator-theory operator-algebras c-star-algebras






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      asked Jan 14 at 13:34









      ervxervx

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          Yes, this is true.



          As $theta$ is rational, it follows that $A_{theta}$ is Morita equivalent to $C(mathbb{T}^2)$, see for example The classification of rational rotation algebras- Brabanter.



          However, two unital $C^*$-algebras are Morita equivalent if and only if each is a full corner of $ntimes n$ matrices over the other, for some $n$, see $C^*$-algebras associated with irrational rotations- Rieffel, Proposition 2.1.



          Thus $A_{m/n}cong pM_k(C(mathbb{T}^2))p$, for some $kin mathbb{N}$ and full prjection $pin M_k(C(mathbb{T}^2))$.






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            1 Answer
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            $begingroup$

            Yes, this is true.



            As $theta$ is rational, it follows that $A_{theta}$ is Morita equivalent to $C(mathbb{T}^2)$, see for example The classification of rational rotation algebras- Brabanter.



            However, two unital $C^*$-algebras are Morita equivalent if and only if each is a full corner of $ntimes n$ matrices over the other, for some $n$, see $C^*$-algebras associated with irrational rotations- Rieffel, Proposition 2.1.



            Thus $A_{m/n}cong pM_k(C(mathbb{T}^2))p$, for some $kin mathbb{N}$ and full prjection $pin M_k(C(mathbb{T}^2))$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Yes, this is true.



              As $theta$ is rational, it follows that $A_{theta}$ is Morita equivalent to $C(mathbb{T}^2)$, see for example The classification of rational rotation algebras- Brabanter.



              However, two unital $C^*$-algebras are Morita equivalent if and only if each is a full corner of $ntimes n$ matrices over the other, for some $n$, see $C^*$-algebras associated with irrational rotations- Rieffel, Proposition 2.1.



              Thus $A_{m/n}cong pM_k(C(mathbb{T}^2))p$, for some $kin mathbb{N}$ and full prjection $pin M_k(C(mathbb{T}^2))$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Yes, this is true.



                As $theta$ is rational, it follows that $A_{theta}$ is Morita equivalent to $C(mathbb{T}^2)$, see for example The classification of rational rotation algebras- Brabanter.



                However, two unital $C^*$-algebras are Morita equivalent if and only if each is a full corner of $ntimes n$ matrices over the other, for some $n$, see $C^*$-algebras associated with irrational rotations- Rieffel, Proposition 2.1.



                Thus $A_{m/n}cong pM_k(C(mathbb{T}^2))p$, for some $kin mathbb{N}$ and full prjection $pin M_k(C(mathbb{T}^2))$.






                share|cite|improve this answer









                $endgroup$



                Yes, this is true.



                As $theta$ is rational, it follows that $A_{theta}$ is Morita equivalent to $C(mathbb{T}^2)$, see for example The classification of rational rotation algebras- Brabanter.



                However, two unital $C^*$-algebras are Morita equivalent if and only if each is a full corner of $ntimes n$ matrices over the other, for some $n$, see $C^*$-algebras associated with irrational rotations- Rieffel, Proposition 2.1.



                Thus $A_{m/n}cong pM_k(C(mathbb{T}^2))p$, for some $kin mathbb{N}$ and full prjection $pin M_k(C(mathbb{T}^2))$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 19 at 13:38









                Shirly GeffenShirly Geffen

                1,316614




                1,316614






























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